Asked by Char
                If cos(t)=(-7/8) where pi<t<(3pi/2), find the values of the following trig functions.
cos(2t)=?
sin(2t)=?
cos(t/2)=?
sin(t/2)=?
            
        cos(2t)=?
sin(2t)=?
cos(t/2)=?
sin(t/2)=?
Answers
                    Answered by
            Reiny
            
    if cos t = -7/8, then sin t = -√15/8 , t is in III
cos 2t = cos^2t - sin^2t = 49/64 - 15/64 = 34/64
sin 2t = 2sintcost = 2(-7/8)(-√15/8) = 14√15/64
cos t = 2cos^2 (t/2) - 1
-7/8 + 1 = 2cos^2 t
cos^2(t/2) = 1/16
cos (t/2) = -1/4 , if t is in III
cos t = 1 - 2sin^2 (t/2)
2sin^2 (t/2) = 1 + 7/8
sin^2 (t/2) = 15/16
sin (t/2) = √15/4 , for t in III
    
cos 2t = cos^2t - sin^2t = 49/64 - 15/64 = 34/64
sin 2t = 2sintcost = 2(-7/8)(-√15/8) = 14√15/64
cos t = 2cos^2 (t/2) - 1
-7/8 + 1 = 2cos^2 t
cos^2(t/2) = 1/16
cos (t/2) = -1/4 , if t is in III
cos t = 1 - 2sin^2 (t/2)
2sin^2 (t/2) = 1 + 7/8
sin^2 (t/2) = 15/16
sin (t/2) = √15/4 , for t in III
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